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Chapter 34: Q4Q (page 1037)

A penguin waddles along the central axis of a concave mirror, from the focal point to an effectively infinite distance. (a) How does its image move? (b) Does the height of its image increase continuously, decrease continuously, or change in some more complicated manner?

Short Answer

Expert verified

(a) The penguin鈥檚 image moves from infinity to the focal point.

(b) The height of the penguin鈥檚 image decreases continuously.

Step by step solution

01

Given data

A penguin waddles along the central axis of a concave mirror from the focal point to an effectively infinite distance.

02

Understanding the concept of optical image

Using the characteristics of the image formed due to a concave mirror when an object is placed at a focal point and an infinite distance from the focal point, we can answer the above questions.

03

(a) Calculation of the reason of movement of the image

The penguin鈥檚 image moves from infinity to the focal point since it waddles along the central axis of a concave mirror, from the focal point to an effectively infinite distance.

04

(b) Calculation of the change in height of the image

The height of the penguin鈥檚 image decreases continuously since it waddles along the central axis of a concave mirror, from the focal point to an effectively infinite distance.

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Most popular questions from this chapter

Figure 34-47a shows the basic structure of the human eye. Light refracts into the eye through the cornea and is then further redirected by a lens whose shape (and thus ability to focus the light) is controlled by muscles. We can treat the cornea and eye lens as a single effective thin lens (Fig. 34-47b). A 鈥渘ormal鈥� eye can focus parallel light rays from a distant object O to a point on the retina at the back of the eye, where the processing of the visual information begins. As an object is brought close to the eye, however, the muscles must change the shape of the lens so that rays form an inverted real image on the retina (Fig. 34-47c). (a) Suppose that for the parallel rays of Figs. 34-47a and b, the focal length f鈥�of the effective thin lens of the eye is 2.50 cm. For an object at distance p = 40 cm, what focal length f鈥� of the effective lens is required for the object to be seen clearly? (b) Must the eye muscles increase or decrease the radii of curvature of the eye lens to give focal length f鈥�?

95 through 100. Three-lens systems. In Fig. 34-49, stick figure O (the object) stands on the common central axis of three thin, symmetric lenses, which are mounted in the boxed regions. Lens 1 is mounted within the boxed region closest to O, which is at object distance p1. Lens 2 is mounted within the middle boxed region, at distance d12 from lens 1. Lens 3 is mounted in the farthest boxed region, at distance d23 from lens 2. Each problem in Table 34-10 refers to a different combination of lenses and different values for distances, which are given in centimeters. The type of lens is indicated by C for converging and D for diverging; the number after C or D is the distance between a lens and either of the focal points (the proper sign of the focal distance is not indicated). Find (a) the image distance i3 for the (final) image produced by lens 3 (the final image produced by the system) and (b) the overall lateral magnification M for the system, including signs. Also, determine whether the final image is (c) real (R) or virtual (V), (d) inverted (I) from object O or non-inverted (NI), and (e) on the same side of lens 3 as object O or on the opposite side.

32 through 38 37, 38 33, 35 Spherical refracting surfaces. An object Ostands on the central axis of a sphericalrefractingsurface. For this situation, each problem in Table 34-5 refers to the index of refractionn1where the object is located, (a) the index of refraction n2on the other side of the refracting surface, (b) the object distance p, (c) the radius of curvature rof the surface, and (d) the image distance i. (All distances are in centimeters.) Fill in the missing information, including whether the image is (e) real (R)or virtual (V)and (f) on the same side of the surface as object Oor on the opposite side

The equation 1p+1i=2rfor spherical mirrors is an approximation that is valid if the image is formed by rays that make only small angles with the central axis. In reality, many of the angles are large, which smears the image a little. You can determine how much. Refer to Fig. 34-22 and consider a ray that leaves a point source (the object) on the central axis and that makes an angle a with that axis. First, find the point of intersection of the ray with the mirror. If the coordinates of this intersection point are x and y and the origin is placed at the center of curvature, then y=(x+p-r)tan a and x2+ y2= r2where pis the object distance and r is the mirror鈥檚 radius of curvature. Next, use tan尾=yxto find the angle b at the point of intersection, and then use伪+y=2尾tofind the value of g. Finally, use the relationtany=y(x+i-r)to find the distance iof the image. (a) Suppose r=12cmand r=12cm. For each of the following values of a, find the position of the image 鈥� that is, the position of the point where the reflected ray crosses the central axis:(0.500,0.100,0.0100rad). Compare the results with those obtained with theequation1p+1i=2r.(b) Repeat the calculations for p=4.00cm.

58 through 67 61 59 Lenses with given radii. Objectstands in front of a thin lens, on the central axis. For this situation, each problem in Table 34-7 gives object distance, index of refraction n of the lens, radiusof the nearer lens surface, and radius of the farther lens surface. (All distances are in centimetres.) Find (a) the image distanceand (b) the lateral magnificationof the object, including signs. Also, determine whether the image is (c) realor virtual, (d) invertedfrom object or non-inverted, and (e) on the same side of the lens as objector on the opposite side.
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